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Finding coefficients for a power series solution for a differential equation (Trig)

The question and my work are attached. I'm having trouble doing the trig parts since I am trying to get all the sums to have an x^n term so can I build my recurrence relation but I'm stuck at getting the trig stuff to have the x^n term.

Could someone please show me what to do?

Any help would be greatly appreciated!

Thanks in advance!

Re: Finding coefficients for a power series solution for a differential equation (Tri

The solution of the second order DE...

$\displaystyle y^{''}-\sin x\ y = \cos x,\ y(0)=-5,\ y^{'}(0)=-9$ (1)

... is supposed to be analytic in x=0 so that we can write...

$\displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}=\sum_{n=0}^{\infty} a_{n}\ x^{n}$ (2)

The computation of the derivatives in x=0 is performed as follows. The initial conditions give us $\displaystyle a_{0}=-5$ and $\displaystyle a_{1}=-9$. From (1) we obtain directly...

$\displaystyle y^{''}= \sin x\ y + \cos x \implies y^{''}(0)= 1 \implies a_{2}=\frac{1}{2}$ (3)

Deriving (3) we obtain...

$\displaystyle y^{'''}= \sin x\ y^{'} + \cos x\ y - \sin x \implies y^{'''}(0)= -5 \implies a_{3}=-\frac{5}{6}$ (4)

The computation of the successive derivatives is left to You...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$